Question

find the derivative of y with respect to z.
y = ln (sinh 2z)
y = \square

Explanation:

Step1: Identify the outer and inner functions

The function \( y = \ln(\sinh 2z) \) is a composite function. Let the outer function be \( f(u) = \ln(u) \) and the inner function be \( u = g(z) = \sinh 2z \).

Step2: Differentiate the outer function

The derivative of \( f(u) = \ln(u) \) with respect to \( u \) is \( f'(u) = \frac{1}{u} \).

Step3: Differentiate the inner function

Now, we need to differentiate \( u = \sinh 2z \) with respect to \( z \). Let \( v = 2z \), so \( u = \sinh(v) \). First, differentiate \( u \) with respect to \( v \): \( \frac{du}{dv} = \cosh(v) \). Then, differentiate \( v \) with respect to \( z \): \( \frac{dv}{dz} = 2 \). By the chain rule, \( \frac{du}{dz} = \frac{du}{dv} \cdot \frac{dv}{dz} = \cosh(v) \cdot 2 = 2\cosh(2z) \) (since \( v = 2z \)).

Step4: Apply the chain rule to the original function

The chain rule states that \( \frac{dy}{dz} = f'(u) \cdot \frac{du}{dz} \). Substituting \( f'(u) = \frac{1}{u} \) (where \( u = \sinh 2z \)) and \( \frac{du}{dz} = 2\cosh(2z) \), we get:

$$ \frac{dy}{dz} = \frac{1}{\sinh 2z} \cdot 2\cosh(2z) $$

Recall the hyperbolic identity \( \coth(x) = \frac{\cosh(x)}{\sinh(x)} \), so \( \frac{\cosh(2z)}{\sinh(2z)} = \coth(2z) \). Therefore, \( \frac{dy}{dz} = 2\coth(2z) \).

Answer:

\( 2\coth(2z) \)